0191. Number of 1 Bits (E)

Number of 1 Bits (E)

题目

Write a function that takes an unsigned integer and return the number of '1' bits it has (also known as the Hamming weight).

Example 1:

Input: 00000000000000000000000000001011
Output: 3
Explanation: The input binary string 00000000000000000000000000001011 has a total of three '1' bits.

Example 2:

Input: 00000000000000000000000010000000
Output: 1
Explanation: The input binary string 00000000000000000000000010000000 has a total of one '1' bit.

Example 3:

Input: 11111111111111111111111111111101
Output: 31
Explanation: The input binary string 11111111111111111111111111111101 has a total of thirty one '1' bits.

Note:

• Note that in some languages such as Java, there is no unsigned integer type. In this case, the input will be given as signed integer type and should not affect your implementation, as the internal binary representation of the integer is the same whether it is signed or unsigned.
• In Java, the compiler represents the signed integers using 2's complement notation. Therefore, in Example 3 above the input represents the signed integer -3.

Follow up:

If this function is called many times, how would you optimize it?

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题意

统计一个32位无符号整数二进制形式中'1'的个数。

思路

最简单的是循环32次，判断每一位上是不是1并统计。

官方解答提供了一种比较巧妙的方法：当n不为0时，说明仍有二进制位为1，迭代时更新 n = n & (n - 1)，这么做可以将n中最后一个1翻转为0。与上一种方法相比，当n为0时即可结束循环，不需要走完32次。

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代码实现

Java

位运算 1

public class Solution {
    // you need to treat n as an unsigned value
    public int hammingWeight(int n) {
        int count = 0;
        for (int i = 0; i < 32; i++) {
            count += (n & 1 << i) != 0 ? 1 : 0;
        }
        return count;
    }
}

位运算 2

public class Solution {
    // you need to treat n as an unsigned value
    public int hammingWeight(int n) {
        int count = 0;
        while (n != 0) {
            count++;
            n &= (n - 1);
        }
        return count;
    }
}

JavaScript

/**
 * @param {number} n - a positive integer
 * @return {number}
 */
var hammingWeight = function (n) {
  let count = 0
  while (n) {
    count++
    n = n & (n - 1)
  }
  return count
}